Martin has severe myopia, with a far point on only 17 cm. He wants to get glasses that he'll wear while using his computer whose screen is 65 cm away. What refractive power will these glasses require?

Respuesta :

Answer:

Explanation:

Far point = 17 cm . That means he can not see beyond this distance .

He wants to see at an object at 65 cm away . That means object placed at 65 has image at 17 cm by concave lens . Using lens formula

1 / v - 1 / u = 1 / f

1 / - 17 - 1 / - 65 = 1 / f

= 1 / 65 - 1 / 17

= -  .0434 = 1 / f

power = - 100 / f

= - 100 x .0434

= - 4.34 D .

Refractive power is the measure of degree of convergence by a lens. The required refractive power for the given glasses is -4. 34 D.

Using lens formula  

[tex]\bold { \dfrac 1 v - \dfrac1 u = \dfrac {1}f}[/tex]

Where,

f-  focal point

v - distance of the image

u - distance of the object  

So,

[tex]\bold { \dfrac 1 {-17} - \dfrac1 {-65} = \dfrac {1}f}\\\\\bold { 0.434 = \dfrac {1}f}\\[/tex]

Since, [tex]\bold {power = \dfrac {- 100 }f}[/tex]

So,

[tex]\bold { power = - 100 \times 0.0434}}\\\\\bold { power = - 4.34\ D}[/tex]  

Therefore, the required refractive power for the given glasses is -4. 34 D.

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